Root of degree n: basic definitions. Root of degree n: basic definitions Removing the minus sign from under the root sign

Congratulations: today we will look at roots - one of the most mind-blowing topics in 8th grade. :)

Many people get confused about roots not because they are complex (what’s so complicated about it - a couple of definitions and a couple more properties), but because in most school textbooks roots are defined through such a jungle that only the authors of the textbooks themselves can understand this writing. And even then only with a bottle of good whiskey. :)

Therefore, now I will give the most correct and most competent definition of a root - the only one that you really should remember. And then I’ll explain: why all this is needed and how to apply it in practice.

But first, remember one important point that many textbook compilers for some reason “forget”:

Roots can be of even degree (our favorite $\sqrt(a)$, as well as all sorts of $\sqrt(a)$ and even $\sqrt(a)$) and odd degree (all sorts of $\sqrt(a)$, $\ sqrt(a)$, etc.). And the definition of a root of an odd degree is somewhat different from an even one.

Probably 95% of all errors and misunderstandings associated with roots are hidden in this fucking “somewhat different”. So let's clear up the terminology once and for all:

Definition. Even root n from the number $a$ is any non-negative the number $b$ is such that $((b)^(n))=a$. And the odd root of the same number $a$ is generally any number $b$ for which the same equality holds: $((b)^(n))=a$.

In any case, the root is denoted like this:

\(a)\]

The number $n$ in such a notation is called the root exponent, and the number $a$ is called the radical expression. In particular, for $n=2$ we get our “favorite” square root (by the way, this is a root of even degree), and for $n=3$ we get a cubic root (odd degree), which is also often found in problems and equations.

Examples. Classic examples of square roots:

\[\begin(align) & \sqrt(4)=2; \\ & \sqrt(81)=9; \\ & \sqrt(256)=16. \\ \end(align)\]

By the way, $\sqrt(0)=0$, and $\sqrt(1)=1$. This is quite logical, since $((0)^(2))=0$ and $((1)^(2))=1$.

Cube roots are also common - no need to be afraid of them:

\[\begin(align) & \sqrt(27)=3; \\ & \sqrt(-64)=-4; \\ & \sqrt(343)=7. \\ \end(align)\]

Well, a couple of “exotic examples”:

\[\begin(align) & \sqrt(81)=3; \\ & \sqrt(-32)=-2. \\ \end(align)\]

If you don’t understand what the difference is between an even and an odd degree, re-read the definition again. This is very important!

In the meantime, we will consider one unpleasant feature of roots, because of which we needed to introduce a separate definition for even and odd exponents.

Why are roots needed at all?

After reading the definition, many students will ask: “What were the mathematicians smoking when they came up with this?” And really: why are all these roots needed at all?

To answer this question, let's go back to elementary school for a moment. Remember: in those distant times, when the trees were greener and the dumplings tastier, our main concern was to multiply numbers correctly. Well, something like “five by five – twenty-five”, that’s all. But you can multiply numbers not in pairs, but in triplets, quadruples and generally whole sets:

\[\begin(align) & 5\cdot 5=25; \\ & 5\cdot 5\cdot 5=125; \\ & 5\cdot 5\cdot 5\cdot 5=625; \\ & 5\cdot 5\cdot 5\cdot 5\cdot 5=3125; \\ & 5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5=15\ 625. \end(align)\]

However, this is not the point. The trick is different: mathematicians are lazy people, so they had a hard time writing down the multiplication of ten fives like this:

That's why they came up with degrees. Why not write the number of factors as a superscript instead of a long string? Something like this:

It's very convenient! All calculations are reduced significantly, and you don’t have to waste a bunch of sheets of parchment and notebooks to write down some 5,183. This record was called a power of a number; a bunch of properties were found in it, but the happiness turned out to be short-lived.

After a grandiose drinking party, which was organized just for the “discovery” of degrees, some particularly stubborn mathematician suddenly asked: “What if we know the degree of a number, but the number itself is unknown?” Now, indeed, if we know that a certain number $b$, say, to the 5th power gives 243, then how can we guess what the number $b$ itself is equal to?

This problem turned out to be much more global than it might seem at first glance. Because it turned out that for most “ready-made” powers there are no such “initial” numbers. Judge for yourself:

\[\begin(align) & ((b)^(3))=27\Rightarrow b=3\cdot 3\cdot 3\Rightarrow b=3; \\ & ((b)^(3))=64\Rightarrow b=4\cdot 4\cdot 4\Rightarrow b=4. \\ \end(align)\]

What if $((b)^(3))=$50? It turns out that we need to find a certain number that, when multiplied by itself three times, will give us 50. But what is this number? It is clearly greater than 3, since 3 3 = 27< 50. С тем же успехом оно меньше 4, поскольку 4 3 = 64 >50. That is this number lies somewhere between three and four, but you won’t understand what it is equal to.

This is precisely why mathematicians came up with $n$th roots. This is precisely why the radical symbol $\sqrt(*)$ was introduced. To designate the very number $b$, which to the indicated degree will give us a previously known value

\[\sqrt[n](a)=b\Rightarrow ((b)^(n))=a\]

I don’t argue: often these roots are easily calculated - we saw several such examples above. But still, in most cases, if you think of an arbitrary number and then try to extract the root of an arbitrary degree from it, you will be in for a terrible bummer.

What is there! Even the simplest and most familiar $\sqrt(2)$ cannot be represented in our usual form - as an integer or a fraction. And if you enter this number into a calculator, you will see this:

\[\sqrt(2)=1.414213562...\]

As you can see, after the decimal point there is an endless sequence of numbers that do not obey any logic. You can, of course, round this number to quickly compare with other numbers. For example:

\[\sqrt(2)=1.4142...\approx 1.4 \lt 1.5\]

Or here's another example:

\[\sqrt(3)=1.73205...\approx 1.7 \gt 1.5\]

But all these roundings, firstly, are quite rough; and secondly, you also need to be able to work with approximate values, otherwise you can catch a bunch of non-obvious errors (by the way, the skill of comparison and rounding is required to be tested on the profile Unified State Examination).

Therefore, in serious mathematics you cannot do without roots - they are the same equal representatives of the set of all real numbers $\mathbb(R)$, just like the fractions and integers that have long been familiar to us.

The inability to represent a root as a fraction of the form $\frac(p)(q)$ means that this root is not a rational number. Such numbers are called irrational, and they cannot be accurately represented except with the help of a radical or other constructions specially designed for this (logarithms, powers, limits, etc.). But more on that another time.

Let's consider several examples where, after all the calculations, irrational numbers will still remain in the answer.

\[\begin(align) & \sqrt(2+\sqrt(27))=\sqrt(2+3)=\sqrt(5)\approx 2.236... \\ & \sqrt(\sqrt(-32 ))=\sqrt(-2)\approx -1.2599... \\ \end(align)\]

Naturally, from the appearance of the root it is almost impossible to guess what numbers will come after the decimal point. However, you can count on a calculator, but even the most advanced date calculator only gives us the first few digits of an irrational number. Therefore, it is much more correct to write the answers in the form $\sqrt(5)$ and $\sqrt(-2)$.

This is exactly why they were invented. To conveniently record answers.

Why are two definitions needed?

The attentive reader has probably already noticed that all the square roots given in the examples are taken from positive numbers. Well, at least from scratch. But cube roots can be calmly extracted from absolutely any number - be it positive or negative.

Why is this happening? Take a look at the graph of the function $y=((x)^(2))$:

The graph of a quadratic function gives two roots: positive and negative

Let's try to calculate $\sqrt(4)$ using this graph. To do this, a horizontal line $y=4$ is drawn on the graph (marked in red), which intersects with the parabola at two points: $((x)_(1))=2$ and $((x)_(2)) =-2$. This is quite logical, since

Everything is clear with the first number - it is positive, so it is the root:

But then what to do with the second point? Like four has two roots at once? After all, if we square the number −2, we also get 4. Why not write $\sqrt(4)=-2$ then? And why do teachers look at such posts as if they want to eat you? :)

The trouble is that if you don’t impose any additional conditions, then the quad will have two square roots - positive and negative. And any positive number will also have two of them. But negative numbers will have no roots at all - this can be seen from the same graph, since the parabola never falls below the axis y, i.e. does not accept negative values.

A similar problem occurs for all roots with an even exponent:

1. Strictly speaking, each positive number will have two roots with even exponent $n$;
2. From negative numbers, the root with even $n$ is not extracted at all.

That is why the definition of an even root of $n$ specifically stipulates that the answer must be a non-negative number. This is how we get rid of ambiguity.

But for odd $n$ there is no such problem. To see this, let's look at the graph of the function $y=((x)^(3))$:

A cube parabola can take any value, so the cube root can be taken from any number

Two conclusions can be drawn from this graph:

1. The branches of a cubic parabola, unlike a regular one, go to infinity in both directions - both up and down. Therefore, no matter what height we draw a horizontal line, this line will certainly intersect with our graph. Consequently, the cube root can always be extracted from absolutely any number;
2. In addition, such an intersection will always be unique, so you don’t need to think about which number is considered the “correct” root and which one to ignore. That is why determining roots for an odd degree is simpler than for an even degree (there is no requirement for non-negativity).

It's a pity that these simple things are not explained in most textbooks. Instead, our brains begin to soar with all sorts of arithmetic roots and their properties.

Yes, I don’t argue: you also need to know what an arithmetic root is. And I will talk about this in detail in a separate lesson. Today we will also talk about it, because without it all thoughts about roots of $n$-th multiplicity would be incomplete.

But first you need to clearly understand the definition that I gave above. Otherwise, due to the abundance of terms, such a mess will begin in your head that in the end you will not understand anything at all.

All you need to do is understand the difference between even and odd indicators. Therefore, let’s once again collect everything you really need to know about roots:

1. A root of an even degree exists only from a non-negative number and is itself always a non-negative number. For negative numbers such a root is undefined.
2. But the root of an odd degree exists from any number and can itself be any number: for positive numbers it is positive, and for negative numbers, as the cap hints, it is negative.

Is it difficult? No, it's not difficult. It's clear? Yes, it’s completely obvious! So now we will practice a little with the calculations.

Basic properties and limitations

Roots have many strange properties and limitations - this will be discussed in a separate lesson. Therefore, now we will consider only the most important “trick”, which applies only to roots with an even index. Let's write this property as a formula:

\[\sqrt(((x)^(2n)))=\left| x\right|\]

In other words, if we raise a number to an even power and then take the root of the same power from it, we will not get the original number, but its modulus. This is a simple theorem that can be easily proven (it is enough to consider non-negative $x$ separately, and then negative ones separately). Teachers constantly talk about it, it is given in every school textbook. But as soon as it comes to solving irrational equations (i.e., equations containing a radical sign), students unanimously forget this formula.

To understand the issue in detail, let’s forget all the formulas for a minute and try to calculate two numbers straight ahead:

\[\sqrt(((3)^(4)))=?\quad \sqrt(((\left(-3 \right))^(4)))=?\]

These are very simple examples. Most people will solve the first example, but many people get stuck on the second. To solve any such crap without problems, always consider the procedure:

1. First, the number is raised to the fourth power. Well, it's kind of easy. You will get a new number that can be found even in the multiplication table;
2. And now from this new number it is necessary to extract the fourth root. Those. no “reduction” of roots and powers occurs - these are sequential actions.

Let's look at the first expression: $\sqrt(((3)^(4)))$. Obviously, you first need to calculate the expression under the root:

\[((3)^(4))=3\cdot 3\cdot 3\cdot 3=81\]

Then we extract the fourth root of the number 81:

Now let's do the same with the second expression. First, we raise the number −3 to the fourth power, which requires multiplying it by itself 4 times:

\[((\left(-3 \right))^(4))=\left(-3 \right)\cdot \left(-3 \right)\cdot \left(-3 \right)\cdot \ left(-3 \right)=81\]

We got a positive number, since the total number of minuses in the product is 4, and they will all cancel each other out (after all, a minus for a minus gives a plus). Then we extract the root again:

In principle, this line could not have been written, since it’s a no brainer that the answer would be the same. Those. an even root of the same even power “burns” the minuses, and in this sense the result is indistinguishable from a regular module:

\[\begin(align) & \sqrt(((3)^(4)))=\left| 3 \right|=3; \\ & \sqrt(((\left(-3 \right))^(4)))=\left| -3 \right|=3. \\ \end(align)\]

These calculations are in good agreement with the definition of a root of an even degree: the result is always non-negative, and the radical sign also always contains a non-negative number. Otherwise, the root is undefined.

Note on procedure

1. The notation $\sqrt(((a)^(2)))$ means that we first square the number $a$ and then take the square root of the resulting value. Therefore, we can be sure that there is always a non-negative number under the root sign, since $((a)^(2))\ge 0$ in any case;
2. But the notation $((\left(\sqrt(a) \right))^(2))$, on the contrary, means that we first take the root of a certain number $a$ and only then square the result. Therefore, the number $a$ can in no case be negative - this is a mandatory requirement included in the definition.

Thus, in no case should one thoughtlessly reduce roots and degrees, thereby allegedly “simplifying” the original expression. Because if the root has a negative number and its exponent is even, we get a bunch of problems.

However, all these problems are relevant only for even indicators.

Removing the minus sign from under the root sign

Naturally, roots with odd exponents also have their own feature, which in principle does not exist with even ones. Namely:

\[\sqrt(-a)=-\sqrt(a)\]

In short, you can remove the minus from under the sign of roots of odd degree. This is a very useful property that allows you to “throw out” all the disadvantages:

\[\begin(align) & \sqrt(-8)=-\sqrt(8)=-2; \\ & \sqrt(-27)\cdot \sqrt(-32)=-\sqrt(27)\cdot \left(-\sqrt(32) \right)= \\ & =\sqrt(27)\cdot \sqrt(32)= \\ & =3\cdot 2=6. \end(align)\]

This simple property greatly simplifies many calculations. Now you don’t need to worry: what if a negative expression was hidden under the root, but the degree at the root turned out to be even? It is enough just to “throw out” all the minuses outside the roots, after which they can be multiplied by each other, divided, and generally do many suspicious things, which in the case of “classical” roots are guaranteed to lead us to an error.

And here another definition comes onto the scene - the same one with which most schools begin the study of irrational expressions. And without which our discussions would be incomplete. Meet!

Arithmetic root

Let's assume for a moment that under the root sign there can only be positive numbers or, in extreme cases, zero. Let's forget about even/odd indicators, let's forget about all the definitions given above - we will work only with non-negative numbers. What then?

And then we will get an arithmetic root - it partially overlaps with our “standard” definitions, but still differs from them.

Definition. An arithmetic root of the $n$th degree of a non-negative number $a$ is a non-negative number $b$ such that $((b)^(n))=a$.

As we can see, we are no longer interested in parity. Instead, a new restriction appeared: the radical expression is now always non-negative, and the root itself is also non-negative.

To better understand how the arithmetic root differs from the usual one, take a look at the graphs of the square and cubic parabola we are already familiar with:

Arithmetic root search area - non-negative numbers

As you can see, from now on we are only interested in those pieces of graphs that are located in the first coordinate quarter - where the coordinates $x$ and $y$ are positive (or at least zero). You no longer need to look at the indicator to understand whether we have the right to put a negative number under the root or not. Because negative numbers are no longer considered in principle.

You may ask: “Well, why do we need such a neutered definition?” Or: “Why can’t we get by with the standard definition given above?”

Well, I will give just one property because of which the new definition becomes appropriate. For example, the rule for exponentiation:

\[\sqrt[n](a)=\sqrt(((a)^(k)))\]

Please note: we can raise the radical expression to any power and at the same time multiply the root exponent by the same power - and the result will be the same number! Here are examples:

\[\begin(align) & \sqrt(5)=\sqrt(((5)^(2)))=\sqrt(25) \\ & \sqrt(2)=\sqrt(((2)^ (4)))=\sqrt(16)\\ \end(align)\]

So what's the big deal? Why couldn't we do this before? Here's why. Let's consider a simple expression: $\sqrt(-2)$ - this number is quite normal in our classical understanding, but absolutely unacceptable from the point of view of the arithmetic root. Let's try to convert it:

$\begin(align) & \sqrt(-2)=-\sqrt(2)=-\sqrt(((2)^(2)))=-\sqrt(4) \lt 0; \\ & \sqrt(-2)=\sqrt(((\left(-2 \right))^(2)))=\sqrt(4) \gt 0. \\ \end(align)$

As you can see, in the first case we removed the minus from under the radical (we have every right, since the exponent is odd), and in the second case we used the above formula. Those. From a mathematical point of view, everything is done according to the rules.

WTF?! How can the same number be both positive and negative? No way. It’s just that the formula for exponentiation, which works great for positive numbers and zero, begins to produce complete heresy in the case of negative numbers.

It was in order to get rid of such ambiguity that arithmetic roots were invented. A separate large lesson is devoted to them, where we consider all their properties in detail. So we won’t dwell on them now - the lesson has already turned out to be too long.

Algebraic root: for those who want to know more

I thought for a long time whether to put this topic in a separate paragraph or not. In the end I decided to leave it here. This material is intended for those who want to understand the roots even better - no longer at the average “school” level, but at one close to the Olympiad level.

So: in addition to the “classical” definition of the $n$th root of a number and the associated division into even and odd exponents, there is a more “adult” definition that does not depend at all on parity and other subtleties. This is called an algebraic root.

Definition. The algebraic $n$th root of any $a$ is the set of all numbers $b$ such that $((b)^(n))=a$. There is no established designation for such roots, so we’ll just put a dash on top:

\[\overline(\sqrt[n](a))=\left\( b\left| b\in \mathbb(R);((b)^(n))=a \right. \right\) \]

The fundamental difference from the standard definition given at the beginning of the lesson is that an algebraic root is not a specific number, but a set. And since we work with real numbers, this set comes in only three types:

1. Empty set. Occurs when you need to find an algebraic root of an even degree from a negative number;
2. A set consisting of one single element. All roots of odd powers, as well as roots of even powers of zero, fall into this category;
3. Finally, the set can include two numbers - the same $((x)_(1))$ and $((x)_(2))=-((x)_(1))$ that we saw on the graph quadratic function. Accordingly, such an arrangement is possible only when extracting the root of an even degree from a positive number.

The last case deserves more detailed consideration. Let's count a couple of examples to understand the difference.

Example. Evaluate the expressions:

\[\overline(\sqrt(4));\quad \overline(\sqrt(-27));\quad \overline(\sqrt(-16)).\]

Solution. With the first expression everything is simple:

\[\overline(\sqrt(4))=\left\( 2;-2 \right\)\]

It is two numbers that are part of the set. Because each of them squared gives a four.

\[\overline(\sqrt(-27))=\left\( -3 \right\)\]

Here we see a set consisting of only one number. This is quite logical, since the root exponent is odd.

Finally, the last expression:

\[\overline(\sqrt(-16))=\varnothing \]

We received an empty set. Because there is not a single real number that, when raised to the fourth (i.e., even!) power, will give us the negative number −16.

Final note. Please note: it was not by chance that I noted everywhere that we work with real numbers. Because there are also complex numbers - it is quite possible to calculate $\sqrt(-16)$ there, and many other strange things.

However, complex numbers almost never appear in modern school mathematics courses. They have been removed from most textbooks because our officials consider the topic “too difficult to understand.”

That's all. In the next lesson we will look at all the key properties of roots and finally learn how to simplify irrational expressions. :)

Examples:

\(\sqrt(16)=2\), since \(2^4=16\)
\(\sqrt(-\frac(1)(125))\) \(=\) \(-\frac(1)(5)\) , since \((-\frac(1)(5) )^3\) \(=\) \(-\frac(1)(125)\)

How to calculate the nth root?

To calculate the root of the \(n\)th power, you need to ask yourself the question: what number to the \(n\)th power will be given under the root?

For example. Calculate the \(n\)th root: a)\(\sqrt(16)\); b) \(\sqrt(-64)\); c) \(\sqrt(0.00001)\); d)\(\sqrt(8000)\); e) \(\sqrt(\frac(1)(81))\).

a) What number to the \(4\)th power will give \(16\)? Obviously, \(2\). That's why:

b) What number to the \(3\)th power will give \(-64\)?

\(\sqrt(-64)=-4\)

c) What number to the \(5\)th power will give \(0.00001\)?

\(\sqrt(0.00001)=0.1\)

d) What number to the \(3\)th power will give \(8000\)?

\(\sqrt(8000)=20\)

e) What number to the \(4\)th power will give \(\frac(1)(81)\)?

\(\sqrt(\frac(1)(81))=\frac(1)(3)\)

We looked at the simplest examples with the \(n\)th root. To solve more complex problems with roots of the \(n\)th degree, it is vital to know them.

Example. Calculate:

\(\sqrt 3\cdot \sqrt(-3) \cdot \sqrt(27) \cdot \sqrt(9) -\) \(=\)		At the moment, none of the roots can be calculated. Therefore, we apply the properties of the root of the \(n\)th degree and transform the expression. \(\frac(\sqrt(-64))(\sqrt(2))\)\(=\)\(\sqrt(\frac(-64)(2))\) \(=\)\(\sqrt(-32)\) because \(\frac(\sqrt[n](a))(\sqrt[n](b))\)\(=\)\(\sqrt[n](\frac(a)(b))\)
\(=\sqrt(3)\cdot \sqrt(-3)\cdot \sqrt(27)\cdot \sqrt(9)-\sqrt(-32)=\)		Let's rearrange the factors in the first term so that the square root and the root of the \(n\)th power are next to each other. This will make it easier to apply properties because Most properties of \(n\)th roots only work with roots of the same degree. And let's calculate the 5th root.
\(=\sqrt(3) \cdot \sqrt(27) \cdot \sqrt(-3)\cdot \sqrt(9)-(-5)=\)		Apply the property \(\sqrt[n](a)\cdot \sqrt[n](b)=\sqrt[n](a\cdot b)\) and expand the bracket
\(=\sqrt(81)\cdot \sqrt(-27)+5=\)		Calculate \(\sqrt(81)\) and \(\sqrt(-27)\)
\(=9\cdot(-3)+5 =-27+5=-22\)

Are the nth root and square root related?

In any case, any root of any degree is just a number, albeit written in a form that is unfamiliar to you.

nth root singularity

The root of the \(n\)th degree with odd \(n\) can be extracted from any number, even negative (see examples at the beginning). But if \(n\) is even (\(\sqrt(a)\), \(\sqrt(a)\),\(\sqrt(a)\)…), then such a root can be extracted only if \( a ≥ 0\) (by the way, the same applies to the square root). This is due to the fact that extracting a root is the opposite of raising to a power.

And raising to an even power makes even a negative number positive. Indeed, \((-2)^6=(-2) \cdot (-2) \cdot (-2) \cdot (-2) \cdot (-2) \cdot (-2)=64\). Therefore, we cannot obtain an even power of a negative number under the root. This means that we cannot extract such a root from a negative number.

An odd power does not have such restrictions - a negative number raised to an odd power will remain negative: \((-2)^5=(-2) \cdot (-2) \cdot (-2) \cdot (-2) \ cdot (-2)=-32\). Therefore, under the root of an odd power you can get a negative number. This means that it is also possible to extract it from a negative number.

Chapter one.

Raising by the square of one-term algebraic expressions.

152. Determination of degree. Recall that the product of two identical numbers ahh called the second power (or square) of a number A , product of three identical numbers ahh called the third power (or cube) of the number A ; in general a work n identical numbers ah... ah called n -th power of number A . The action by which the power of a given number is found is called raising to a power (second, third, etc.). The repeating factor is called the base, and the number of identical factors is called the exponent.

The degrees are abbreviated as follows: a 2, a 3, a 4 ...etc.

We will first talk about the simplest case of exponentiation, namely elevation squared; and then we will consider elevation in other degrees.

153. Rule of signs when raising a square. From the rule for multiplying relative numbers it follows that:

(+2) 2 =(+2) (+2) = + 4; (+ 1 / 3) 2 =(+ 1 / 3)(+ 1 / 3) = + 1 / 9 ;

(-2) 2 =(-2) (-2) = + 4; (- 1 / 3) 2 =(- 1 / 3)(- 1 / 3) = + 1 / 9

(+a) 2 =(+a) (+a) = +a 2

(-a) 2 =(-a) (-a) = +a 2

This means that the square of any relative number is a positive number.

154. Squaring products, powers and fractions.

A) Suppose you need to square the product of several factors, for example. abc . This means that it is required abc multiply by abc . But to multiply by the product abc , you can multiply the multiplicand by A , multiply the result by b and what can be multiplied by With .

(abc) 2 = (abc) (abc) = (abc) abc = abcabc

(we have dropped the last parentheses, since this does not change the meaning of the expression). Now, using the associative property of multiplication ( department 1§ 34, b), we group the factors as follows:

(aa) (bb) (ss),

which can be abbreviated as: a 2 b 2 c 2.

Means, to square the product, you can square each factor separately
(To shorten the speech, this rule, like the one that follows, is not fully expressed; it would be necessary to add: “and multiply the results obtained.” The addition of this is self-evident..)

Thus:

(3 / 4 xy) 2 = 9 / 16 x 2 y 2 ; (- 0.5mn) 2 = + 0.25m 2 n 2 ; etc.

b) Let some degree be required, for example. a 3 , raise to square. This can be done like this:

(a 3) 2 = a 3 a 3 = a 3+3 = a 6.

Like this: (x 4) 2 = x 4 x 4 = x 4+4 = x 8

Means, To raise a power by a square, you can multiply the exponent by 2 .

Thus, applying these two rules, we will have, for example:

(- 3 3 / 4 a x 2 y 3) 2 = (- 3 3 / 4) 2 a 2 (x 2) 2 (y 3) 2 = 225 / 2 a 2 x 4 y 6

V) Suppose you want to square some fraction a / b . Then, applying the rule of multiplying a fraction by a fraction, we get:

Means, To square a fraction, you can square the numerator and denominator separately.

Example.

Chapter two.

Squaring a polynomial.

155. Derivation of the formula. Using the formula ( department2 chapter3§ 61):

(a + b) 2 = a 2 + 2ab + b 2 ,

we can square the trinomial a + b + c , treating it as a binomial (a + b) + c :

(a + b + c) 2 = [(a + b) + c ] 2 = (a + b) 2 + 2(a + b)c + c 2 = a 2 + 2ab + b 2 + 2(a + b)c + c 2

Thus, by adding to the binomial a + b third member With after the elevation, 2 terms were added to the square: 1) double the product of the sum of the first two terms by the third term and 2) the square of the third term. Let us now apply to the trinomial a + b + c another fourth member d and raise the fourfold a + b + c + d squared, taking the sum a + b + c for one member.

(a + b +c + d) 2 = [(a + b + c) + d ] 2 = (a + b +c) 2 + 2(a + b + c)d + d 2

Substituting instead (a + b +c) 2 we find the expression we obtained above:

(a + b +c + d) 2 = a 2 + 2ab + b 2 + 2(a + b)c + c 2 + 2(a + b + c)d + d 2

We again notice that with the addition of a new term, 2 terms are added to the raised polynomial in its square: 1) double the product of the sum of the previous terms and the new term and 2) the square of the new term. Obviously, this addition of two terms will continue as new terms are added to the elevated polynomial. Means:

The square of a polynomial is equal to: the square of the 1st term, plus twice the product of the 1st term by the 2nd, plus the square of the 2nd term, plus double the product of the sum of the first two terms by the 3rd, plus the square of the 3rd term, plus double the product of the sum of the first three terms and the 4th term, plus the square of the 4th term, etc. Of course, the terms of a polynomial can also be negative.

156. A note about signs. The final result with a plus sign will be, firstly, the squares of all terms of the polynomial and, secondly, those doubled products that resulted from multiplying terms with the same signs.

Example.

157. Reduced elevation to the square of integers. Using the formula for the square of a polynomial, you can square any integer differently than by ordinary multiplication. Let, for example, you want to square 86 . Let's break this number down into digits:

86 = 80 + 6 = 8 dec. + 6 units.

Now, using the formula for the square of the sum of two numbers, we can write:

(8 dec. + 6 units) 2 = (8 dec.) 2 + 2(8 dec.) (6 units) + (6 units) 2 .

To quickly calculate this sum, consider that the square of tens is hundreds (but can be thousands); eg 8 des.. squared form 64 hundreds, because 80 2 = b400; the product of tens and ones is tens (but can also be hundreds), e.g. 3 dec. 5 units = 15 des, since 30 5 = 150; and the square of units is ones (but can be tens), e.g. 9 units squared = 81 units. Therefore, it is most convenient to arrange the calculation as follows:

that is, we first write the square of the first digit (hundreds); under this number we write twice the product of the first digit by the second (tens), observing that the last digit of this product is one place to the right of the last digit of the top number; then, again moving the last digit one place to the right, place the square of the second digit (one); and add all the written numbers into one sum. Of course, one could supplement these numbers with the appropriate number of zeros, i.e. write it like this:

but this is useless, if only we correctly sign the numbers under each other, each time retreating (with the last digit) one place to the right.

Let it also be required to square 238 . Because:

238 = 2 cells. + 3 dec. + 8 units, That

But hundreds squared give tens of thousands (for example, 5 hundred squared will be 25 tenth thousand, since 500 2 = 250,000), the product of hundreds times tens gives thousands (for example, 500 30 = 15,000), etc. .

Examples.

Chapter three.

y = x 2 And y = ah 2 .

158. Graph of a function y = x 2 . Let's see how when changing the number raised X its square changes X 2 (for example, how when the side of a square changes, its area changes). To do this, let us first pay attention to the following features of the function y = x 2 .

A) For any value X the function is always possible and always receives only one specific value. For example, when X = - 10 the function will be (-10) 2 = 100 , at
X =1000 the function will be 1000 2 =1 000 000 , etc.

b) Because (- X ) 2 = X 2 , then for two values X , differing only in signs, two identical positive values ​​are obtained at ; for example, when X = - 2 and at X = + 2 meaning at it will be the same thing, exactly 4 . Negative values ​​for at it never works out.

V) If the absolute value of x increases without limit, then at increases indefinitely. So, if for X we will give a series of unlimitedly increasing positive values: 1, 2, 3, 4... or a series of unlimitedly decreasing negative values: -1, -2, -3, -4..., then for at we get a series of unlimitedly increasing values: 1, 4, 9, 16, 25... These briefly express, saying that when x = + ∞ and at x = - ∞ function at done + ∞ .

G) X at . So, if the value x = 2 , let's give an increment, let's put 0,1 (i.e. instead of x = 2 let's take x = 2.1 ), That at instead of 2 2 = 4 will become equal

(2 + 0,1) 2 = = 2 2 + 2 2 0,1 + 0,1 2 .

Means, at will increase by 2 2 0,1 + 0,1 2 = 0,41 . If the same value X let's give an even smaller increment, let's say 0,01 , then y becomes equal

(2 + 0,01) 2 = = 2 2 + 2 2 0,01 + 0,01 2 . .

This means that then y will increase by 2 2 0,01 + 0,01 2 = 0,0401 , i.e. it will increase less than before. In general, the smaller the fraction we increase X , the smaller the number will increase at . Thus, if we imagine that X increases (let us assume from the value 2) continuously, passing through all values ​​greater than 2, then at will also increase continuously, passing through all values ​​greater than 4.

Having noticed all these properties, let's create a table of function values y = x 2 , for example, like this:

Let us now depict these values ​​in the drawing in the form of points, the abscissa of which will be the written values X , and the ordinates are the corresponding values at (in the drawing we took a centimeter as a unit of length); We outline the resulting points with a curve. This curve is called a parabola.

Let's look at some of its properties.

A) A parabola is a continuous curve, since with a continuous change in the abscissa X (both in a positive and negative direction) the ordinate, as we have seen now, also changes continuously.

b) The entire curve is located on one side of the axis x -ov, exactly on the side on which the positive ordinate values ​​lie.

V) A parabola is subdivided by an axis at -s into two parts (branches). Dot ABOUT , at which these branches converge is called the vertex of the parabola. This point is the only common point between the parabola and the axis x -s; this means that at this point the parabola touches the axis x -s.

G) Both branches are infinite, since X And at can increase indefinitely. Branches rise from the axis x -s unlimitedly upward, moving at the same time unlimitedly from the axis y -s right and left.

d) Axis y -ov serves as the axis of symmetry for the parabola, so that by bending the drawing along this axis so that the left half of the drawing falls on the right, we will see that both branches will align; for example, a point with abscissa - 2 and ordinate 4 will be compatible with a point with abscissa +2 and the same ordinate 4.

e) At X = 0 the ordinate is also equal to 0. This means that when X = 0 function has the smallest possible value. The function does not have the greatest significance, since the ordinates of the curve increase indefinitely.

159. Graph of a function of the formy = ah 2 . Let us first assume that A is a positive number. Let's take, for example, these 2 functions:

1) y= 1 1 / 2 x 2 ; 2) y= 1 / 3 x 2

Let's make tables of the values ​​of these functions, for example, like this:

Let's plot all these values ​​on the drawing and draw the curves. For comparison, we placed another graph of the function on the same drawing (dashed line):

3) y=x 2

From the drawing it is clear that with the same abscissa the ordinate of the 1st curve in 1 1 / 2 , times larger, and the ordinate of the 2nd curve is 3 times less than the ordinate of the 3rd curve. As a result, all such curves have a common character: infinite continuous branches, axis of symmetry, etc., only when a > 1 the branches of the curve are more elevated, and when a< 1 they are more bent downwards than those of the curve y=x 2 . All such curves are called parabolammy.

Let us now assume that the coefficient A will be a negative number. Let, for example, y = - 1 / 3 x 2 . Comparing this function with this one: y = + 1 / 3 x 2 we notice that for the same value X both functions have the same absolute value, but are opposite in sign. Therefore, in the drawing for the function y = - 1 / 3 x 2 you get the same parabola as for the function y= 1 / 3 x 2 only located under the axis X -ov symmetrically with a parabola y= 1 / 3 x 2 . In this case, all values ​​of the function are negative, except for one, which is equal to zero at x = 0 ; this last value is the largest of all.

Comment. If the relationship between two variables at And X is expressed by the equality: y = ah 2 , Where A some constant number, then we can say that the quantity at proportional to the square of the quantity X , since with an increase or decrease X 2 times, 3 times, etc. value at increases or decreases by 4 times, 9 times, 16 times, etc. For example, the area of ​​a circle is π R 2 , Where R is the radius of the circle and π constant number (equal to approximately 3.14); Therefore, we can say that the area of ​​a circle is proportional to the square of its radius.

Chapter Four.

Raising to cube and other powers of one-term algebraic expressions.

160. Rule of signs when raising to a degree. From the rule for multiplying relative numbers it follows that

(-5) 3 = (-5)(-5)(-5) = -125;

(- 1 / 2 ) 4 = (- 1 / 2 ) (- 1 / 2 ) (- 1 / 2 ) (- 1 / 2 )=+ 1 / 16 ;

(- 1) 5 = (- 1) (- 1) (- l) (-1) (-1) = - l;

(- 1) 6 = (- 1) (- 1) (- l) (-1) (-1) (-1) = +l; etc.

Means, Raising a negative number to a power with an even exponent produces a positive number, and raising it to a power with an odd exponent produces a negative number.

161. Raising to the power of products, powers and fractions. When raising the product of a degree and a fraction to some power, we can do the same as when raising it by a square (). So:

(abc) 3 = (abc)(abc)(abc) = abc abc abc = (aaa)(bbb)(ccs) = a 3 b 3 c 3 ;

Chapter Five.

Graphic representation of functions: y = x 3 and y = ax 3 .

162. Graph of a function y = x 3 . Let's consider how, when the number being raised changes, its cube changes (for example, how when the edge of a cube changes, its volume changes). To do this, we first indicate the following features of the function y = x 3 (reminiscent of the properties of the function y = x 2 , considered by us earlier):

A) For any value X function y = x 3 possible and has only one meaning; Thus, (+ 5) 3 = +125 and the cube of + 5 cannot be equal to any other number. Similarly, (- 0.1) 3 = - 0.001 and the cube of the number -0.1 cannot be equal to any other number.

b) With two values X , differing only in signs, function x 3 receives values ​​that also differ from each other only in signs; yes, when X = 2 function x 3 equal to 8, and when X = - 2 it is equal - 8 .

V) As x increases, the function x 3 is increasing, and moreover, faster than X , and even faster than x 2 ; so when

X = - 2, -1, 0, +1, + 2, +3, + 4. .. x 3 will = -8, - 1, 0, +1, + 8, +27, + 64 ...

G) Very small increment of variable number X corresponds to a very small increment of the function x 3 . So, if the value X = 2 increase by a fraction 0,01 , i.e. if instead X = 2 let's take x = 2,01 , then the function at there will be no 2 3 (i.e. not 8 ), A 2,01 3 , which will be 8,120601 . This means that this function will then increase by 0,120601 . If the value X = 2 let's increase it even less, for example, by 0,001 , That x 3 will become equal 2,001 3 , which will be 8,012006001 , and, therefore, at will only increase by 0,012006001 . We see, therefore, that if the increment of a variable number X will be less and less, then the increment x 3 there will be less and less.

Noticing this property of the function y = x 3 , let's draw a graph of it. To do this, first compile a table of the values ​​of this function, for example, like this:

163. Graph of a function y = ax 3 . Let's take these two functions:

1) y = 1 / 2 x 3 ; 2) y = 2 x 3

If we compare these functions with a simpler one: y = x 3 , then we note that for the same value X the first function receives values ​​half as large, and the second twice as large as the function y = ax 3 , in all other respects these three functions are similar to each other. Their graphs are shown for comparison in the same drawing. These curves are called parabolas of the 3rd degree.

Chapter six.

Basic properties of root extraction.

164. Tasks.

A) Find the side of a square whose area would be equal to the area of ​​a rectangle with a base of 16 cm and a height of 4 cm.

Marking the side of the required square with a letter X (cm), we get the following equation:

x 2 =16 4, i.e. x 2 = 64.

We see in this way that X there is a number which, when raised to the second power, gives the result 64. Such a number is called the second root of 64. It is equal to + 8 or - 8, since (+ 8) 2 = 64 and (- 8) 2 = 64. The negative number - 8 is not suitable for our task, since the side of the square must be expressed as an ordinary arithmetic number.

b) A piece of lead weighing 1 kg 375 g (1375 g) is cube-shaped. How big is the edge of this cube if we know that 1 cubic. cm of lead weighs 11 grams?

Let the length of the edge of the cube be X cm. Then its volume will be equal x 3 cube cm, and his weight will be 11 x 3 G.

11x 3= 1375; x 3 = 1375: 11 = 125.

We see in this way that X there is a number which, when raised to the third power, is 125 . This number is called third root out of 125. It is, as you might guess, equal to 5, since 5 3 = 5 5 5 = 125. This means that the edge of the cube referred to in the problem has a length of 5 cm.

165. Definition of root. The second degree (or square) root of a number A is the number whose square is equal to A . Thus, the square root of 49 is 7, and also - 7, since 7 2 = 49 and (- 7) 2 = 49. The third degree (cubic) root of the number A is the number whose cube equals A . So, the cube root of -125 is -5, since (- 5) 3 = (-5)(-5)(-5)= -125.

Generally the root n-th degree from among A is the number that n the th power is equal to A.

Number n , meaning what degree the root is, is called root index.

The root is denoted by the sign √ (radical sign, i.e. root sign). Latin word radix means root. Sign√ first introduced in the 15th century.. Under the horizontal line they write the number from which the root is found (radical number), and above the hole in the angle they put the index of the root. So:

The cube root of 27 is denoted..... 3 √27 ;

The fourth root of 32 is denoted... 3 √32.

It is customary not to write the square root exponent at all, for example.

instead of 2 √16 they write √16.

The action by which the root is found is called root extraction; it is the opposite of elevation to a degree, since through this action one finds what is given during elevation to a degree, namely the basis of the wall, and what is given is what is found during elevation to a degree, namely the degree itself. Therefore, we can always verify the correctness of extracting the root by raising it to a degree. For example, to check

equality: 3 √125 = 5, it is enough to cube 5: having received the radical number 125, we conclude that the cube root of 125 was extracted correctly.

166. Arithmetic root. A root is called an arithmetic root if it is taken from a positive number and is itself a positive number. For example, the arithmetic square root of 49 is 7, whereas the number - 7, which is also the square root of 49, cannot be called arithmetic.

Let us indicate the following two properties of the arithmetic root.

a) Let it be necessary to find the arithmetic √49. Such a root will be 7, since 7 2 = 49. Let us ask ourselves whether it is possible to find some other positive number X , which would also be √49. Let's assume that such a number exists. Then it must be either less than 7 or greater than 7. If we assume that x < 7, то тогда и x 2 < 49 (с уменьшением множимого и множителя произведение уменьшается); если же допустим, что x >7, then x 2 >49. This means that no positive number, neither less than 7 nor greater than 7, can be equal to √49. Thus, there can be only one arithmetic root of a given degree from a given number.

We would come to a different conclusion if we were talking not about the positive meaning of the root, but about some kind; Thus, √49 is equal to both the number 7 and the number - 7, since both 7 2 = 49 and (- 7) 2 = 49.

b) Let's take some two unequal positive numbers, for example. 49 and 56. From the fact that 49< 56, мы можем заключить, что и √49 < √56 (если только знаком √ будем обозначать арифметический квадратный корень). Действительно: 7 < 8. Подобно этому из того, что 64 < l25, мы можем заключить, что и 3 √64 < 3 √125

Indeed: 3 √64 = 4 and 3 √125 = 5 and 4< 5. Вообще a smaller positive number also corresponds to a smaller arithmetic root (same degree).

167. Algebraic root. A root is called algebraic if it is not required that it be taken from a positive number and that it itself be positive. Thus, if under the expression n √a of course the algebraic root n th degree, then this means that the number A It can be both positive and negative, and the root itself can be both positive and negative.

Let us indicate the following 4 properties of an algebraic root.

A) An odd root of a positive number is a positive number .

So, 3 √8 must be a positive number (it is 2), since a negative number raised to a power with an odd exponent produces a negative number.

b) An odd root of a negative number is a negative number.

So, 3 √-8 must be a negative number (it is -2), since a positive number raised to any power produces a positive number, not a negative one.

V) An even root of a positive number has two values ​​with opposite signs and the same absolute value.

Yes, √ +4 = + 2 and √ +4 = - 2 , because (+ 2 ) 2 = + 4 And (- 2 ) 2 = + 4 ; exactly the same 4 √+81 = + 3 And 4 √+81 = - 3 , because both degrees (+3) 4 And (-3) 4 equal to the same number. The double value of a root is usually indicated by placing two signs in front of the absolute value of the root; they write like this:

√4 = ± 2 ; √a 2 = ± a ;

G) An even root of a negative number cannot be equal to any positive or negative number. , since both of them, when raised to a power with an even exponent, give a positive number, not a negative one. For example, √ -9 is not equal to +3, or -3, or any other number.

An even root of a negative number is usually called an imaginary number; relative numbers are called real, or valid, numbers.

168. Extracting the root of a product, a degree and a fraction.

A) Suppose we need to extract the square root of the product abc . If it were necessary to square the product, then, as we saw (), it is possible to square each factor separately. Since extracting a root is the inverse action of raising a power, we must expect that to extract a root from a product one can extract it from each factor separately, i.e. that

√abc = √a √b √c .

To verify the correctness of this equality, let’s raise the right side of it by a square (according to the theorem: to raise a product to a power...):

(√a √b √c ) 2 = (√a ) 2 (√b ) 2 (√c ) 2

But according to the definition of a root,

(√a ) 2 = a, (√b ) 2 = b, (√c ) 2 = c

Hence

(√a √b √c ) 2 = abc .

If the square of the product √ a √b √c equals abc , then this means that the product is equal to the square root of abc .

Like this:

3 √abc = 3 √a 3 √b 3 √c ,

(3 √a 3 √b 3 √c ) 3 = (3 √a ) 3 (3 √b ) 3 (3 √c ) 3 = abc

Means, To extract a root from a product, it is enough to extract it from each factor separately.

b) It is easy to verify that the following equalities are true:

√a 4 = A 2 , because (a 2 ) 2 = A 4 ;

3 √x 12 = x 4 , „ (x 4 ) 3 = x 12 ; etc.

Means, To extract the root of a power whose exponent is divisible by the exponent of the root, you can divide the exponent by the exponent of the root.

V) The following equalities will also be true:

Means, To extract the root of a fraction, you can use the numerator and denominator separately.

Note that these truths assume that we are talking about arithmetic roots.

Examples.

1) √9a 4 b 6 = √9 √a 4 √b 6 = 3A 2 b 3 ;

2) 3 √125a 6 x 9 = 3 √125 3 √a 6 3 √x 9 = 5A 2 x 3

Note If the required root is of even degree and is assumed to be algebraic, then a double sign must be placed in front of the result found: ± Thus,

√9x 4 = ± 3x 2 .

169. The simplest transformations of radicals,

A) Taking factors beyond the radical sign. If the radical expression is decomposed into factors such that the root can be extracted from some of them, then such factors, after extracting the root from them, can be written before the radical sign (can be taken out behind the radical sign).

1) √a 3 = √a 2 a = √a 2 √a = A √a .

2) √24 a 4 x 3 = √4 6 a 4 x 2 x = 2a 2 x √6x

3) 3 √16 x 4 = 3 √8 2 x 3 x = 2x 3 √2 x

b) Subsuming factors under the radical sign. Sometimes it is useful, on the contrary, to subsume the factors that precede it under the sign of the radical; To do this, it is enough to raise such factors to a power whose exponent is equal to the exponent of the radical, and then write the factors under the radical sign.

Examples.

1) A 2 √a = √(A 2 ) 2 a = √A 4 a = √a 5 .

2) 2x 3 √x = 3 √(2x ) 3 x = 3 √8x 3 x = 3 √8x 4 .

V) Freeing radical expressions from denominators. Let's show this with the following examples:

1) Transform the fraction so that the square root can be extracted from the denominator. To do this, multiply both terms of the fraction by 5:

2) Multiply both terms of the fraction by 2 , on A and on X , i.e. on 2Oh :

Comment. If you need to extract the root of an algebraic sum, then it would be a mistake to extract it from each term separately. Eg.√ 9 + 16 = √25 = 5 , whereas
√9 + √16 = 3 + 4 = 7 ; means the action of rooting in relation to addition (and subtraction) does not have distributive properties(as well as exaltation, department 2 chapter 3§ 61, remark).